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Magazine Chapter 11: Problem 30
Find the cube roots of each complex number. Leave the answers in trigonometric form. Then graph each cube root as a vector in the complex plane. $$\sqrt{3}-i$$
Short Answer
Expert verified
The cube roots of \( \sqrt{3} - i \) are \( 2^{1/3}(\cos(-\pi/18) + i\sin(-\pi/18)) \), \( 2^{1/3}(\cos(11\pi/18) + i\sin(11\pi/18)) \), and \( 2^{1/3}(\cos(23\pi/18) + i\sin(23\pi/18)) \).
Step by step solution
01
Express in Polar Form
First, we write the complex number \( \sqrt{3} - i \) in polar form, which is \( r(\cos \theta + i\sin \theta) \). We calculate the magnitude \( r \) as \( r = \sqrt{ (\sqrt{3})^2 + (-1)^2 } = \sqrt{3 + 1} = 2 \). The angle \( \theta \) is found using \( \tan^{-1}\left(\frac{-1}{\sqrt{3}}\right) \), which simplifies to \( \theta = -\frac{\pi}{6} \). So, the polar form is: \( 2 \left( \cos\left(-\frac{\pi}{6}\right) + i\sin\left(-\frac{\pi}{6}\right) \right) \).
02
Calculate Cube Roots
To find the cube roots, we use the formula for the roots of a complex number in polar form: \[z_k = r^{1/n} \left( \cos\left(\frac{\theta + 2k\pi}{n}\right) + i\sin\left(\frac{\theta + 2k\pi}{n}\right) \right)\], where \( n = 3 \). Here, \( r^{1/3} = 2^{1/3} \) and we solve for \( k = 0, 1, 2 \).
03
Finding the First Root (k = 0)
For \( k = 0 \), \( z_0 = 2^{1/3} \left( \cos\left( \frac{-\pi/6}{3} \right) + i \sin\left( \frac{-\pi/6}{3} \right) \right) \). Simplifying gives \( z_0 = 2^{1/3} \left( \cos\left( -\frac{\pi}{18} \right) + i \sin\left( -\frac{\pi}{18} \right) \right) \).
04
Finding the Second Root (k = 1)
For \( k = 1 \), \( z_1 = 2^{1/3} \left( \cos\left( \frac{-\pi/6 + 2\pi}{3} \right) + i \sin\left( \frac{-\pi/6 + 2\pi}{3} \right) \right) \). This simplifies to \( z_1 = 2^{1/3} \left( \cos\left( \frac{11\pi}{18} \right) + i \sin\left( \frac{11\pi}{18} \right) \right) \).
05
Finding the Third Root (k = 2)
For \( k = 2 \), \( z_2 = 2^{1/3} \left( \cos\left( \frac{-\pi/6 + 4\pi}{3} \right) + i \sin\left( \frac{-\pi/6 + 4\pi}{3} \right) \right) \). Simplifying gives \( z_2 = 2^{1/3} \left( \cos\left( \frac{23\pi}{18} \right) + i \sin\left( \frac{23\pi}{18} \right) \right) \).
06
Graphing the Roots
On the complex plane, each root \( z_0, z_1, z_2 \) is drawn as a vector from the origin. These vectors represent the angles \( -\frac{\pi}{18} \), \( \frac{11\pi}{18} \), and \( \frac{23\pi}{18} \) respectively, each scaled by the modulus, \( 2^{1/3} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Cube Roots
Cube roots of a complex number are the three distinct values that, when cubed, result in the original number. To find these roots, we use a method involving polar and trigonometric forms of complex numbers. For the complex number \( \sqrt{3} - i \), we first convert it into polar form, then apply the formula for cube roots in polar coordinates. Here are the steps you will follow:
- Compute the magnitude \( r \) of the complex number \((\sqrt{3} - i)\). For our number, \( r = 2 \).
- Determine the angle \( \theta \) in its polar form. Here, \( \theta = -\frac{\pi}{6} \).
- Use the formula \[ z_k = r^{1/3} \left( \cos\left(\frac{\theta + 2k\pi}{3}\right) + i\sin\left(\frac{\theta + 2k\pi}{3}\right) \right) \] to find each of the three roots. Calculate for \( k = 0, 1, \text{and } 2 \).
Polar Form
Complex numbers can be expressed in polar form, which simplifies many operations, such as multiplication, division, and finding roots. The polar form of a complex number is given by \( r(\cos \theta + i\sin \theta) \). Here's what you need to understand about it:
- \( r \) is the magnitude or modulus of the complex number, calculated as \( r = \sqrt{x^2 + y^2} \), where \( x \) and \( y \) are the real and imaginary parts, respectively.
- \( \theta \) is the argument or angle of the complex number, determined by \( \tan^{-1}\left(\frac{y}{x}\right) \).
Trigonometric Form
Closely related to polar form, the trigonometric form of a complex number uses the same concepts but focuses more on the expressions involving sine and cosine functions. It is written as \( r(\cos \theta + i\sin \theta) \). Here's how you can interpret it:
- The trigonometric form highlights the geometric nature of complex numbers, showcasing how they behave on the complex plane as vectors.
- This form is particularly helpful in representing periodic functions, which makes it crucial for solving roots like cube roots of complex numbers, maintaining the angle increments precisely.
- By breaking complex numbers into cosine and sine components, operations like exponentiation and finding roots become more convenient.
Complex Plane
The complex plane is a two-dimensional plane used to visualize complex numbers graphically. Also known as the Argand plane, it enables us to depict any complex number as a vector from the origin. Here is how you can understand its application:
- Complex numbers have both real and imaginary components. The real part is represented on the horizontal axis, while the imaginary part is shown on the vertical axis.
- Operations such as addition or multiplication are visualized as geometric transformations, like translations or rotations, respectively.
- Finding roots, like cube roots, generates multiple vectors that are symmetrically placed around a circle centered at the origin. For example, when graphing the cube roots of a number, you get three vectors forming a perfect equilateral arrangement based on the calculated angles.
- This graphical representation helps in understanding relationships and transformations involving complex operations.
